一道英语物理题A quarterback passes the football downfield at 20m/s .It leaves his hand 1.8m above the ground and is caught by a receiver 30m away at the same heightWhat is the maximum height of the ball on its way to the receiver?

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一道英语物理题Aquarterbackpassesthefootballdownfieldat20m/s.Itleaveshishand1.8mabovethegroundandiscaughtbya

一道英语物理题A quarterback passes the football downfield at 20m/s .It leaves his hand 1.8m above the ground and is caught by a receiver 30m away at the same heightWhat is the maximum height of the ball on its way to the receiver?
一道英语物理题
A quarterback passes the football downfield at 20m/s .It leaves his hand 1.8m above the ground and is caught by a receiver 30m away at the same height
What is the maximum height of the ball on its way to the receiver?

一道英语物理题A quarterback passes the football downfield at 20m/s .It leaves his hand 1.8m above the ground and is caught by a receiver 30m away at the same heightWhat is the maximum height of the ball on its way to the receiver?
翻译如下：
一个球员以20m/s的速度抛出了一只足球,球离开他手时离地1.8m,接着,球在同一高度下被30米外的另一个球员接住了
请问,球的运动轨迹上可能的最大高度为多少?
要设角度来做的,设抛出时速度方向和水平夹角为A
所以球的水平速度是20cosA,竖直速度是20sinA,方向向上
所以球运动的时间就是t=s/v=30/20cosA
所以竖直速度就要满足20sinA/g=0.5*30/20cosA
算出角度A为0.5arcsin0.75,得出水平速度,竖直速度,运动时间
所以最大高度就为0.5*竖直速度*运动时间=3.386m

关于速度力的

计算抛物线最高点的。
大致意思是，一个橄榄球以20m/s的速度在距离地面1.8米的高度被投出，如果在30m外，能够在同样的高度接住，问这个橄榄球在飞行过程中的最大高度是多少

说白了就是初速度为20m/s的物体做抛体运动，水平位移30m，垂直为0.列方程解就行了，水平初速度20cos x，垂直20sin x ，最后解得垂直位移为10*（1-根号7再除以4）约为3.39m。
真够累的，这是哪里出的数学题啊……汗颜,不会是你想知道qb传给跑峰的球能否在空中拦截？呵呵，放心，3.3米多呢，一般人是断不了的...

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